Disclaimer: I dont mean that I've discovered a conceptually completely different way of proving those laws, of course. I just found myself proving them like this and then realized that they're presented differently in my sources, so I'm wondering whether my presentations are valid, because they makes more sense intuitively to me.
A) Proof that $\log_b xy = \log_bx + \log_by$:
Let $x = b^a$ and $y = b^c$
Then, $\log_bxy = \log_b b^ab^c = \log_bb^{a+c} = a+c = \log_bb^a + \log_bb^c = log_bx+log_by$
My question here is whether it's ok for me to assume that I can always find $b^a$ to represent $x$ and $b^c$ to represent $y$ (is there a proof of this, by the way?).
B) Proof that $\log_bx^n = n \log_bx$
Here I wanted to prove this law using the first law, as such:
$\log_bx^n = \log_b(x_1 \cdot x_2 \cdot ... \cdot x_n) = \log_bx_1 + \log_bx_2 + ... + \log_bx_n = n \cdot \log_bx$
Is this a valid proof? It makes sense to me but it seems a step is missing between the last two to make it airtight.
EDIT: Beyond just validity, I welcome any criticism of elegantness, formatting, etc..
On the face of it, these are fine.
In the first, supposing that there exists $a$ such that $b^a = x$ is the same as saying the exponential function $b^{(\cdot)}$ surjects onto the positive reals (assuming that $x > 0$ throughout). We can rigorously prove this. One way is to show that $b^x$ as a function of $x$ is continuous and $\lim_{x \to -\infty} b^x = 0$ and $\lim_{x \to \infty} b^x = \infty$. Then the intermediate value theorem gives you what you need.
The other clear method of showing it involves showing that the domain of the log function contains all positive numbers, and that the logarithm is the inverse function of the corresponding exponentiation.
Which you prefer might depend on what your definitions of log and exponentials are.
In the second, it's a bit unfortunate that your displayed proof only works for integer $n$, when it holds in much greater generality. You shouldn't use $x_1, x_2, \ldots, x_n$ to refer to $n$ copies of $x$, since they're all the same $x$. If you want to highlight that there are $n$ copies, you might use $$ \log_b x^n = \log_b (\underbrace{x \cdot x \cdots x}_{n \text{ copies}} ).$$